Răspuns :
a+b=24
(a-3)=(b+3)x1/5
a=(b+3)x1/5+3
(b+3)x1/5+3+b=24
(b+3)x1/5+b=21 | x5
b+3+5b=105
6b=102
b=102/6
b=17
a=24-17
a=7
Ana=7 nuci
Bianca=17 nuci
(a-3)=(b+3)x1/5
a=(b+3)x1/5+3
(b+3)x1/5+3+b=24
(b+3)x1/5+b=21 | x5
b+3+5b=105
6b=102
b=102/6
b=17
a=24-17
a=7
Ana=7 nuci
Bianca=17 nuci
notam cu a-nr nucilor anei
b-nr nucilor biancai
a+b=24=>
a=b+3 unde a=[tex] \frac{1}{5} [/tex] * b=>
1.b+3+b=25
b(3+1)=24
b*4=24
b= 6
atunci a = [tex] \frac{1}{5} [/tex] *6 =[tex] \frac{6}{5} [/tex]
b-nr nucilor biancai
a+b=24=>
a=b+3 unde a=[tex] \frac{1}{5} [/tex] * b=>
1.b+3+b=25
b(3+1)=24
b*4=24
b= 6
atunci a = [tex] \frac{1}{5} [/tex] *6 =[tex] \frac{6}{5} [/tex]